Skip to content

HTTPS clone URL

Subversion checkout URL

You can clone with
or
.
Download ZIP
Fetching contributors…

Cannot retrieve contributors at this time

405 lines (325 sloc) 12.909 kB
"""
Copyright (C) 2010 David Fong and Michael Saunders
LSMR uses an iterative method.
07 Jun 2010: Documentation updated
03 Jun 2010: First release version in Python
David Chin-lung Fong clfong@stanford.edu
Institute for Computational and Mathematical Engineering
Stanford University
Michael Saunders saunders@stanford.edu
Systems Optimization Laboratory
Dept of MS&E, Stanford University.
"""
from __future__ import division, print_function, absolute_import
__all__ = ['lsmr']
from numpy import zeros, infty
from numpy.linalg import norm
from math import sqrt
from scipy.sparse.linalg.interface import aslinearoperator
from .lsqr import _sym_ortho
def lsmr(A, b, damp=0.0, atol=1e-6, btol=1e-6, conlim=1e8,
maxiter=None, show=False):
"""Iterative solver for least-squares problems.
lsmr solves the system of linear equations ``Ax = b``. If the system
is inconsistent, it solves the least-squares problem ``min ||b - Ax||_2``.
A is a rectangular matrix of dimension m-by-n, where all cases are
allowed: m = n, m > n, or m < n. B is a vector of length m.
The matrix A may be dense or sparse (usually sparse).
.. versionadded:: 0.11.0
Parameters
----------
A : {matrix, sparse matrix, ndarray, LinearOperator}
Matrix A in the linear system.
b : (m,) ndarray
Vector b in the linear system.
damp : float
Damping factor for regularized least-squares. `lsmr` solves
the regularized least-squares problem::
min ||(b) - ( A )x||
||(0) (damp*I) ||_2
where damp is a scalar. If damp is None or 0, the system
is solved without regularization.
atol, btol : float
Stopping tolerances. `lsmr` continues iterations until a
certain backward error estimate is smaller than some quantity
depending on atol and btol. Let ``r = b - Ax`` be the
residual vector for the current approximate solution ``x``.
If ``Ax = b`` seems to be consistent, ``lsmr`` terminates
when ``norm(r) <= atol * norm(A) * norm(x) + btol * norm(b)``.
Otherwise, lsmr terminates when ``norm(A^{T} r) <=
atol * norm(A) * norm(r)``. If both tolerances are 1.0e-6 (say),
the final ``norm(r)`` should be accurate to about 6
digits. (The final x will usually have fewer correct digits,
depending on ``cond(A)`` and the size of LAMBDA.) If `atol`
or `btol` is None, a default value of 1.0e-6 will be used.
Ideally, they should be estimates of the relative error in the
entries of A and B respectively. For example, if the entries
of `A` have 7 correct digits, set atol = 1e-7. This prevents
the algorithm from doing unnecessary work beyond the
uncertainty of the input data.
conlim : float
`lsmr` terminates if an estimate of ``cond(A)`` exceeds
`conlim`. For compatible systems ``Ax = b``, conlim could be
as large as 1.0e+12 (say). For least-squares problems,
`conlim` should be less than 1.0e+8. If `conlim` is None, the
default value is 1e+8. Maximum precision can be obtained by
setting ``atol = btol = conlim = 0``, but the number of
iterations may then be excessive.
maxiter : int
`lsmr` terminates if the number of iterations reaches
`maxiter`. The default is ``maxiter = min(m, n)``. For
ill-conditioned systems, a larger value of `maxiter` may be
needed.
show : bool
Print iterations logs if ``show=True``.
Returns
-------
x : ndarray of float
Least-square solution returned.
istop : int
istop gives the reason for stopping::
istop = 0 means x=0 is a solution.
= 1 means x is an approximate solution to A*x = B,
according to atol and btol.
= 2 means x approximately solves the least-squares problem
according to atol.
= 3 means COND(A) seems to be greater than CONLIM.
= 4 is the same as 1 with atol = btol = eps (machine
precision)
= 5 is the same as 2 with atol = eps.
= 6 is the same as 3 with CONLIM = 1/eps.
= 7 means ITN reached maxiter before the other stopping
conditions were satisfied.
itn : int
Number of iterations used.
normr : float
``norm(b-Ax)``
normar : float
``norm(A^T (b - Ax))``
norma : float
``norm(A)``
conda : float
Condition number of A.
normx : float
``norm(x)``
References
----------
.. [1] D. C.-L. Fong and M. A. Saunders,
"LSMR: An iterative algorithm for sparse least-squares problems",
SIAM J. Sci. Comput., vol. 33, pp. 2950-2971, 2011.
http://arxiv.org/abs/1006.0758
.. [2] LSMR Software, http://www.stanford.edu/~clfong/lsmr.html
"""
A = aslinearoperator(A)
b = b.squeeze()
msg=('The exact solution is x = 0 ',
'Ax - b is small enough, given atol, btol ',
'The least-squares solution is good enough, given atol ',
'The estimate of cond(Abar) has exceeded conlim ',
'Ax - b is small enough for this machine ',
'The least-squares solution is good enough for this machine',
'Cond(Abar) seems to be too large for this machine ',
'The iteration limit has been reached ')
hdg1 = ' itn x(1) norm r norm A''r'
hdg2 = ' compatible LS norm A cond A'
pfreq = 20 # print frequency (for repeating the heading)
pcount = 0 # print counter
m, n = A.shape
# stores the num of singular values
minDim = min([m, n])
if maxiter is None:
maxiter = minDim
if show:
print(' ')
print('LSMR Least-squares solution of Ax = b\n')
print('The matrix A has %8g rows and %8g cols' % (m, n))
print('damp = %20.14e\n' % (damp))
print('atol = %8.2e conlim = %8.2e\n' % (atol, conlim))
print('btol = %8.2e maxiter = %8g\n' % (btol, maxiter))
u = b
beta = norm(u)
v = zeros(n)
alpha = 0
if beta > 0:
u = (1 / beta) * u
v = A.rmatvec(u)
alpha = norm(v)
if alpha > 0:
v = (1 / alpha) * v
# Initialize variables for 1st iteration.
itn = 0
zetabar = alpha * beta
alphabar = alpha
rho = 1
rhobar = 1
cbar = 1
sbar = 0
h = v.copy()
hbar = zeros(n)
x = zeros(n)
# Initialize variables for estimation of ||r||.
betadd = beta
betad = 0
rhodold = 1
tautildeold = 0
thetatilde = 0
zeta = 0
d = 0
# Initialize variables for estimation of ||A|| and cond(A)
normA2 = alpha * alpha
maxrbar = 0
minrbar = 1e+100
normA = sqrt(normA2)
condA = 1
normx = 0
# Items for use in stopping rules.
normb = beta
istop = 0
ctol = 0
if conlim > 0:
ctol = 1 / conlim
normr = beta
# Reverse the order here from the original matlab code because
# there was an error on return when arnorm==0
normar = alpha * beta
if normar == 0:
if show:
print(msg[0])
return x, istop, itn, normr, normar, normA, condA, normx
if show:
print(' ')
print(hdg1, hdg2)
test1 = 1
test2 = alpha / beta
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (normr, normar)
str3 = ' %8.1e %8.1e' % (test1, test2)
print(''.join([str1, str2, str3]))
# Main iteration loop.
while itn < maxiter:
itn = itn + 1
# Perform the next step of the bidiagonalization to obtain the
# next beta, u, alpha, v. These satisfy the relations
# beta*u = a*v - alpha*u,
# alpha*v = A'*u - beta*v.
u = A.matvec(v) - alpha * u
beta = norm(u)
if beta > 0:
u = (1 / beta) * u
v = A.rmatvec(u) - beta * v
alpha = norm(v)
if alpha > 0:
v = (1 / alpha) * v
# At this point, beta = beta_{k+1}, alpha = alpha_{k+1}.
# Construct rotation Qhat_{k,2k+1}.
chat, shat, alphahat = _sym_ortho(alphabar, damp)
# Use a plane rotation (Q_i) to turn B_i to R_i
rhoold = rho
c, s, rho = _sym_ortho(alphahat, beta)
thetanew = s*alpha
alphabar = c*alpha
# Use a plane rotation (Qbar_i) to turn R_i^T to R_i^bar
rhobarold = rhobar
zetaold = zeta
thetabar = sbar * rho
rhotemp = cbar * rho
cbar, sbar, rhobar = _sym_ortho(cbar * rho, thetanew)
zeta = cbar * zetabar
zetabar = - sbar * zetabar
# Update h, h_hat, x.
hbar = h - (thetabar * rho / (rhoold * rhobarold)) * hbar
x = x + (zeta / (rho * rhobar)) * hbar
h = v - (thetanew / rho) * h
# Estimate of ||r||.
# Apply rotation Qhat_{k,2k+1}.
betaacute = chat * betadd
betacheck = -shat * betadd
# Apply rotation Q_{k,k+1}.
betahat = c * betaacute
betadd = -s * betaacute
# Apply rotation Qtilde_{k-1}.
# betad = betad_{k-1} here.
thetatildeold = thetatilde
ctildeold, stildeold, rhotildeold = _sym_ortho(rhodold, thetabar)
thetatilde = stildeold* rhobar
rhodold = ctildeold * rhobar
betad = - stildeold * betad + ctildeold * betahat
# betad = betad_k here.
# rhodold = rhod_k here.
tautildeold = (zetaold - thetatildeold * tautildeold) / rhotildeold
taud = (zeta - thetatilde * tautildeold) / rhodold
d = d + betacheck * betacheck
normr = sqrt(d + (betad - taud)**2 + betadd * betadd)
# Estimate ||A||.
normA2 = normA2 + beta * beta
normA = sqrt(normA2)
normA2 = normA2 + alpha * alpha
# Estimate cond(A).
maxrbar = max(maxrbar, rhobarold)
if itn > 1:
minrbar= min(minrbar, rhobarold)
condA = max(maxrbar, rhotemp) / min(minrbar, rhotemp)
# Test for convergence.
# Compute norms for convergence testing.
normar = abs(zetabar)
normx = norm(x)
# Now use these norms to estimate certain other quantities,
# some of which will be small near a solution.
test1 = normr / normb
if (normA * normr) != 0:
test2 = normar / (normA * normr)
else:
test2 = infty
test3 = 1 / condA
t1 = test1 / (1 + normA * normx / normb)
rtol = btol + atol * normA * normx / normb
# The following tests guard against extremely small values of
# atol, btol or ctol. (The user may have set any or all of
# the parameters atol, btol, conlim to 0.)
# The effect is equivalent to the normAl tests using
# atol = eps, btol = eps, conlim = 1/eps.
if itn >= maxiter:
istop = 7
if 1 + test3 <= 1:
istop = 6
if 1 + test2 <= 1:
istop = 5
if 1 + t1 <= 1:
istop = 4
# Allow for tolerances set by the user.
if test3 <= ctol:
istop = 3
if test2 <= atol:
istop = 2
if test1 <= rtol:
istop = 1
# See if it is time to print something.
if show:
if (n <= 40) or (itn <= 10) or (itn >= maxiter - 10) or \
(itn % 10 == 0) or (test3 <= 1.1 * ctol) or \
(test2 <= 1.1 * atol) or (test1 <= 1.1 * rtol) or \
(istop != 0):
if pcount >= pfreq:
pcount = 0
print(' ')
print(hdg1, hdg2)
pcount = pcount + 1
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (normr, normar)
str3 = ' %8.1e %8.1e' % (test1, test2)
str4 = ' %8.1e %8.1e' % (normA, condA)
print(''.join([str1, str2, str3, str4]))
if istop > 0:
break
# Print the stopping condition.
if show:
print(' ')
print('LSMR finished')
print(msg[istop])
print('istop =%8g normr =%8.1e' % (istop, normr))
print(' normA =%8.1e normAr =%8.1e' % (normA, normar))
print('itn =%8g condA =%8.1e' % (itn, condA))
print(' normx =%8.1e' % (normx))
print(str1, str2)
print(str3, str4)
return x, istop, itn, normr, normar, normA, condA, normx
Jump to Line
Something went wrong with that request. Please try again.